The decomposition of HI(g) is represented by the e...

Question

The decomposition of HI(g) is represented by the equation

2HI(g) ⇌ H₂(g) + I₂(g)

The following experiment was devised to determine the equilibrium constant of the reaction. HI (g) is introduced into five identical 400-cm³ glass bulbs, and the five bulbs are maintained at 623 K. The amount of I₂ produced over time is measured by opening each bulb and titrating the contents with 0.0150 M Na₂S₂O₃ (aq). The reaction of I₂ with the titrant is

I₂ + 2Na₂S₂O₃ ⇌ Na₂S₄O₆ + 2NaI

Data for the experiment are provided in this table.

What is the value of Kc for the decomposition of HI at 623 K?

Jules Bruno · Accepted Answer

We’re asked to determine the equilibrium constant Kc for the decomposition of HI at 623 K, based on the titration data.In equilibrium, the amount of I2 produced is the same amount that was titrated by Na2S2O3.Titration reaction occurs in a 1:2 mole ratio, where 1 mole of I2 reacts with 2 moles of Na2S2O3.This way, we can determine the amount of I2 in the decomposition reaction.To determine Kc, we do the following steps:Step 1: Calculate the moles of I2 in equilibrium, from the volume of titrant used.[readmore]Step 2: Calculate the moles of H2 in equilibrium.Step 3: Calculate the initial moles of HI.Step 4: Calculate the moles of HI in equilibrium.Step 5: Calculate the concentrations of HI, H2, and I2 in equilibrium.Step 6: Calculate Kc.Step 1: Moles of I2 in equilibriumFirst, we determine the Molar Mass (MM) of Na2S2O3:Na2S2O3 2 Na x 23.00 g/mol = 46.00 g/mol                2 S   x 32.07 g/mol = 64.14 g/mol                3 O   x 16.00 g/mol = 48.00 g/mol                                           Sum = 158.14 g/molThen, we calculate the moles of Na2S2O3 for each bulb, given that [Na2S2O3] = 0.0150 M:We rearrange the equation for moles:And calculate the moles of Na2S2O3 in each bulb given that M = 0.0150 M Na2S2O3:Finally, we determine the moles at the equilibrium of I2 from the 1:2 molar ratio:Step 2: Moles of H2 in equilibriumThe decomposition reaction of HI is:2 HI(g) ↔ H2(g) + I2(g)From mole to mole comparison, the moles of H2 and I2 produced are the same:Bulb 1: moles of H2 = moles of I2 = 1.572x10-4 molBulb 2: moles of H2 = 2.092x10-4 molBulb 3: moles of H2 = 2.423x10-4 molBulb 4: moles of H2 = 3.112x10-4 molBulb 5: moles of H2 = 2.151x10-4 molStep 3: Initial moles of HI.First, we determine the MM of HI:HI 1 I   x 126.90 g/mol = 126.90 g/mol     1 H x     1.01 g/mol = 1.01 g/mol                                    Sum = 127.91 g/molThen, we determine the initial moles of HI in each bulb:Moles HI = (mass of HI)/(MM of HI)Step 4: Moles of HI in equilibrium.For all the bulbs, the moles of HI in equilibrium are determined from an ICE chart where we can use either moles or concentrations.                        2 HI(g)     ↔     H2(g)      +     I2(g)Where x = initial moles of H2 and I2.From it, we determine the moles of HI in equilibrium as follows:Moles of HI in equilibrium = initial moles of HI – 2 * moles of product (either I2 or H2)Bulb 1: moles of HI = 2.345x10-3 mol – 2(1.572x10-4 mol) = 2.031x10-3 molBulb 2: moles of HI = 2.502x10-3 mol – 2(2.092x10-4 mol) = 2.084x10-4 molBulb 3: moles of HI = 2.463x10-3 mol – 2(2.423x10-4 mol) = 1.978x10-4 molBulb 4: moles of HI = 3.174x10-3 mol – 2(3.112x10-4 mol) = 2.552x10-4 molBulb 5: moles of HI = 2.189x10-3 mol – 2(2.151x10-4 mol) = 1.759x10-4 molStep 5: Concentrations of HI, H2, and I2 in equilibrium.The concentration is determined as follows:We know the moles of HI, H2, and I2 in equilibrium.Also, the problem states that V = 400 cm3. So first, we convert from cm3 to L considering that 1 cm3 = 1 mL:Then, in each bulb, we calculate the Molarity of HI, H2, and I2. Remember that moles of H2 = moles of I2:Bulb 1:Bulb 2:Bulb 3:Bulb 4:Bulb 5:Step 6: Calculate Kc.For the decomposition reaction of HI:2 HI(g) ↔ H2(g) + I2(g)The Kc expression is:Note that each concentration is raised to the stoichiometric coefficient: [H2] and [I2] are raised to 1 while [HI] is raised to 2.Since [H2] = [I2], the expression becomes:Plugging the equilibrium concentrations determined in Step 5, we determine Kc for each bulb:For the first 3 bulbs, the equilibrium constant Kc increases with increasing time but after that, Kc value is constant. Hence, at 623 K the Kc is 0.0150.

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